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3 years ago

I need help PLEASE!! :O

Physics

2 answers:

Reil [10]3 years ago

5 0

Idk if this with help or not but here ya go !!

svet-max [94.6K]3 years ago

4 0

Answer:

Protons and electrons create electric fields. Most electric charge is carried by the electrons and protons within an atom. Electrons are said to carry negative charge, while protons are said to carry positive charge.

Explanation:

none

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What type of pressure is associated with more dense, sinking air?

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B I hope it helps you

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3 years ago

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In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What

allsm [11]

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

$m=\dfrac{W}{g}$

$m=\dfrac{22.3}{9.8}$

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$k=4\pi^2 f^2m$

$k=4\pi^2 \times (2.7)^2\times 2.27$

k = 653.3 N/m

So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.

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4 years ago

Which type of wave requires a medium?

melamori03 [73]

The correct answer is option C. Sound wave

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3 years ago

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A 100 kg cart goes around the inside of a vertical loop of a roller coaster. The radius of the loop is 3 m and the cart moves at

Inga [223]

Answer:

220N

Explanation:

Weight of the cart = w = 100kg

Radius = r = 3m

Velocity = v = 6m/s

9 = 9.8m/s²

F = ?

Since the cart is at the top of inner loop both the average normal force (N) and the weight (w=mg) of the cart act downwards, hence they both add up. Considering the centrifugal force (F=mv²/r) as well, we get,

∑F=ma=mv²/r

N+mg=mv²/r

N=m(v²/r-g)

N=100[(6)²/3-9.8]

N=100[12-9.8]

Net Force=220N

So, the force exerted by the track on the cart at the top of the loop is 220N

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3 years ago

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A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

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3 years ago